Question

In mendel dihybrid cross, F2 progeny produced 800 seeds. Of these how many resemble F1 hybrid phenotypically and genotypically respectively?
1)400-200
2)450-200
3)400-250
4)450-250
Please answer... ​

Answer 1

Explanation:

dihybrid cross between two pure varieties that differ in two pairs of contrasting characters obtains F

1

generation. Selfing between two hybrids give the phenotypic ratio of 9:3:3:1 in F

2

generation, out of which 9/16 have the phenotype identical to F

1

hybrid. Therefore, out of 128 plants in F

2

generation, a number of the plant which are phenotypically identical to F

1

hybrid would be 9/16 * 128= 72. The genotypic ratio for F

2

generation is 1: 2 :1 : 4 : 2 : 2 :1 : 2 :1ttrr out of which 4/16 have the genotype identical to F

1

hybrid. Therefore, out of 128 plants in F

2

generation, a number of the plant which are genotypically identical to F

1

hybrid would be 4/16 * 128= 32.

Answer 2

Answer:

Explanation:

Hi,

PHENOTYPIC RATIO

For dihybrid cross the f2 progeny givest the phenotypic ratio as

9:3:3:1 and the no of progeny having phenotype as ROUNG YELLOW is 9 (refer attachment -1).

so if we take the ratio we get 9/16*800 = 450 resembles the phenotype of ROUND YELLOW.

GENOTYPIC RATIO

For dihybrid cross the f2 progeny givest the genotypic ratio as

1:2:2: 4 :1:2:1:2:1 where the no of the progeny resembling the hybrid progeny i.e RrYy is 4 as shown in the attachment-2  given.

so if we take the ratio as 4/16 * 800 =200 progeny resembles  the genotypic ratio of RrYy.

So option 2 is correct.

Hope u understood, Thank you

Harini.M

QHM